40 g NaOH. You must use 40 g NaOH to prepare 10.0 L of a solution that has a pH of 13.
<em>Step 1</em>. Calculate the pOH of the solution
pOH = 14.00 – pH = 14.00 -13 = 1
<em>Step 2</em>. Calculate the concentration of NaOH
[NaOH] = [OH^(-)] = 10^(-pOH) mol/L = 10^(-1) mol/L = 0.1 mol/L
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 10.0 L solution × (0.1 mol NaOH/1 L solution) = 1 mol NaOH
<em>Step 4</em>. Calculate the mass of NaOH
Mass of NaOH = 1 mol NaOH × (40.00 g NaOH/1 mol NaOH) = 40 g NaOH
In a flame photometric analysis, salt solution is first vaporized using the heat of flame, followed by this electrons from valance shell gets excited from ground state to excited state. Followed by this de-excitation of electron bring backs electrons to ground state. This process is accompanied by emission of photon. The photon emitted is characteristic of an element, and number of photons emitted can be used for quantitative analysis.
<span>Following are the investigative question that you can answer by doing this experiment.
</span>1) What information can be obtained from the colour of flame?
2) <span>State the relationship between wavelength, frequency, and energy?
</span><span>3) Can you identify the metal present in unknown sample provided?
4) How will you identify amount of metal present in sample solution?
5) </span><span>Why do different chemicals emit light of different colour?</span><span>
</span>
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,
pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
![\frac{[Deprotonated]}{[Protonated]}=0.01](https://tex.z-dn.net/?f=%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D%3D0.01)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
![\frac{[Protonated]}{[Deprotonated]}=100](https://tex.z-dn.net/?f=%5Cfrac%7B%5BProtonated%5D%7D%7B%5BDeprotonated%5D%7D%3D100)
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
Answer:
51.0 meters/second is the final speed
Answer:
Since the earliest stars that formed in the universe (Population III stars) do not have the required metallicity to host planets, it is believed that the supernova explosions from such stars helped enrich subsequent (Population II) stars, some of which may still be in existence and could host planets.
Explanation:
hope this helps
